ĐKXĐ : \(4\le x\le6\)

Xét \(VP^2=6-x+x-4+2\sqrt{\left(6-x\right)\left(x-4\right)}=2+2\sqrt{\left(6-x\right)\left(x-4\right)}\)

Áp dụng bđt Cauchy ta có : \(2+2\sqrt{\left(6-x\right)\left(x-4\right)}\le2+6-x+x-4=4\)

\(\Rightarrow VP\le2\forall x\)(1)

Xét \(VT=x^2-10x+27=\left(x^2-10x+25\right)+2=\left(x-5\right)^2+2\ge2\forall x\)(2)

Từ (1);(2) \(\Rightarrow VT\ge2\ge VP\)

Dấu "=" xảy ra \(\hept{\begin{cases}6-x=x-4\\\left(x-5\right)^2=0\end{cases}\Rightarrow x=5\left(TMĐKXĐ\right)}\)

Vậy nghiệm pt là x = 5

ta có đề bài <=> 

\(\sqrt{\left(x-3\right)^2}+\sqrt{\left(x+5\right)^2}=8\)

<=> \(\left|x-3\right|+\left|x+5\right|=8\)

<=>\(\left|3-x\right|+\left|x+5\right|=8\)

Áp dụng tính chât dấu giá trị tuyệt đối ta có 

\(\left|3-x\right|+\left|x+5\right|>=\left|3-x+x+5\right|=8\)

dấu = xảy ra <=> \(\left(3-x\right)\left(x+5\right)>=0\)

đến đây bạn tự giaỉ dấu = nhé

\(x+y+z-6046=2\sqrt{x-2019}+4\sqrt{y-2020}+6\sqrt{z-2021}\)

\(\left(x-2019\right)+\left(x-2020\right)+\left(x-2021\right)+1+4+9\)\(=2\sqrt{x-2019}+4\sqrt{y-2020}+6\sqrt{z-2021}\)

đặt :\(\hept{\begin{cases}\sqrt{x-2019}=a\\\sqrt{y-2020}=b\\\sqrt{z-2021}=c\end{cases}\left(đk:a,b,c\ge0\right)}\)

PT <=>  \(a^2+b^2+c^2+1+4+9=2a+4b+6c\)

\(\Leftrightarrow\left(a-1\right)^2+\left(b-2\right)^2+\left(c-6\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}a-1=0\\b-2=0\\c-3=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=1\\b=2\\c=3\end{cases}\left(tm\right)}}\)

\(\Rightarrow\hept{\begin{cases}x=2020\\y=2024\\z=2030\end{cases}}\)

a: ĐKXĐ: x>=0

b: \(\Leftrightarrow\dfrac{2\sqrt{2}-2\sqrt{2-\sqrt{x}}+\sqrt{2x}-\sqrt{x\left(2-\sqrt{x}\right)}+2\sqrt{2}+2\sqrt{2+\sqrt{x}}-\sqrt{2x}-\sqrt{x\left(2+\sqrt{x}\right)}}{2-2+\sqrt{x}}=\sqrt{2}\)

\(\Leftrightarrow4\sqrt{2}-2\sqrt{x\left(\sqrt{x}+2\right)}=\sqrt{2x}\)

\(\Leftrightarrow\sqrt{4x\left(\sqrt{x}+2\right)}=4\sqrt{2}-\sqrt{2x}\)

\(\Leftrightarrow4x\left(\sqrt{x}+2\right)=32-16\sqrt{x}+2x\)

\(\Leftrightarrow4x\sqrt{x}+8x-32+16\sqrt{x}-2x=0\)

=>\(x\in\left\{0;1.2996\right\}\)

Áp dụng BĐT:\(\left|A\right|+\left|B\right|\ge\left|A+B\right|\)

Ta có: \(\left|\sqrt{x-1}+2\right|+\left|3-\sqrt{x-1}\right|\ge\left|\sqrt{x-1}+2+3-\sqrt{x-1}\right|=5\)

Dấu \(=\)xảy ra khi \(AB\ge0\)

dat \(\sqrt{x-1}\) = t

ta có: \(\sqrt{x+3+4t}\)+ \(\sqrt{x+8-6t}\)= 5

     x + 3 + 4t + x + 8 - 6t = 25

   2x - 2t = 14 ( chia cả 2 vế cho 2)

   x - t = 7

   t = x - 7

  thay t = \(\sqrt{x}-1\)vào ta được:

 x - 7 = \(\sqrt{x-1}\)

( x - 7 )² = x - 1

x² -14x + 49 = x - 1

x² - 15x + 50 = 0

​k biết đúng hay k

OoO Ledegill2 OoO. Ban co the giai thich ro hon giup minh duoc khong. hi

a, \(\sqrt{2}x-\sqrt{6}=0\Leftrightarrow\sqrt{2}x=\sqrt{6}\Leftrightarrow x=\sqrt{3}\)

b, \(\frac{x^2}{\sqrt{3}}-\sqrt{12}=0\Leftrightarrow\frac{x^2}{\sqrt{3}}=\sqrt{12}\Leftrightarrow x^2=\sqrt{12}.\sqrt{3}\Leftrightarrow x^2=\sqrt{36}\Leftrightarrow x=36\)

c, \(\sqrt{3}x+\sqrt{3}=\sqrt{12}+\sqrt{27}\Leftrightarrow\sqrt{3}x=\sqrt{12}+\sqrt{27}-\sqrt{3}\)

\(\Leftrightarrow\sqrt{3}x=2\sqrt{3}+3\sqrt{3}-\sqrt{3}\Leftrightarrow\sqrt{3}x=4\sqrt{3}\Leftrightarrow x=4\)

a.

\(\sqrt{4x^2+4x+1}-\sqrt{25x^2+10x+1}=0\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}-\sqrt{\left(5x+1\right)^2}=0\)

\(\Leftrightarrow2x+1-\left(5x+1\right)=0\)

\(\Leftrightarrow-3x=0\Leftrightarrow x=0\)

b.

\(\sqrt{x^4-16x^2+64}=\sqrt{25x^2+10x+1}\)

\(\Leftrightarrow\sqrt{\left(x^2-8\right)^2}=\sqrt{\left(5x+1\right)^2}\)

\(\Leftrightarrow x^2-8=5x+1\)

\(\Leftrightarrow x^2-5x+\dfrac{25}{4}=\dfrac{61}{4}\)

\(\Leftrightarrow\left(x-\dfrac{5}{2}\right)^2=\dfrac{61}{4}\)

............................

tương tự ..

c: \(\Leftrightarrow\sqrt{x-5}\left(\sqrt{x+5}-1\right)=0\)

=>x-5=0 hoặc x+5=1

=>x=-4 hoặc x=5

d: \(\Leftrightarrow\sqrt{2x+3}\left(\sqrt{2x-3}-2\right)=0\)

=>2x+3=0 hoặc 2x-3=4

=>x=7/2 hoặc x=-3/2

e: \(\Leftrightarrow\sqrt{x-2}\left(1-3\sqrt{x+2}\right)=0\)

=>x-2=0 hoặc 3 căn x+2=1

=>x=2 hoặc x+2=1/9

=>x=-17/9 hoặc x=2